Multiple choice

The system of equations x + y + z = 6 x + 4y + 6z = 20 x + 4y + $\lambda z$ = $\mu$ has NO solution for values of $\lambda$ and $\mu$ given by

  1. $\lambda$ = 6, $\mu$ = 20
  2. $\lambda$ = 6, $\mu$ $\neq$ 20
  3. $\lambda$ $\neq$ 6, $\mu$ = 20
  4. $\lambda$ $\neq$ 6, $\mu$ $\neq$ 20
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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